# Physical vs. UnPhysical

Gold Member
I was and am reading quite a lot of books on GR and QFT and there's the distinction between a physical state and an unphysical state.
What is the difference?
What makes a state physical?

I know this is a bit philosophical .
Doesn't this distinction change with time and experiments?

## Answers and Replies

Vanadium 50
Staff Emeritus
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I think this will fall into the Potter Stewart "I know it when I see it" definition.

As an example, I might solve for the electric field from a configuration of charges and have two solutions - one that gets small when I am far from the ensemble of charges and one that gets larger the farther away I get. I would dismiss the second solution as unphysical.

dextercioby, vanhees71, Demystifier and 5 others
Mister T
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Gold Member
I was and am reading quite a lot of books on GR and QFT and there's the distinction between a physical state and an unphysical state.
Can you tell us more about which books you're reading and the context in which they're making this distinction?

When I think of a state it implies the existence of a system. That is, a system is in some particular state. If the state is physical then there are experiments that can be performed, at least in principle, to verify the state. On the other hand if there is no experiment that can be performed, even in principle, to verify the state of a system then the state is not physical.

hutchphd
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What makes a state physical?
I think the designation as unphysical is merely one of convenience, and not, in a narrow sense, at all worrisome. Subsystems (this contrivance or that field) are written down without initially worrying about interactions with all the myriad other degrees of freedom. To solve a real world system boundaries in time and space must be specified. It turns out that some subset of boundary conditions are never encountered for the subsystem and these are simply declared inutile. There is no particular reason to expect otherwise.

Gold Member
Can you tell us more about which books you're reading and the context in which they're making this distinction?

When I think of a state it implies the existence of a system. That is, a system is in some particular state. If the state is physical then there are experiments that can be performed, at least in principle, to verify the state. On the other hand if there is no experiment that can be performed, even in principle, to verify the state of a system then the state is not physical.
Specifically, now I am reading on and off the book Perutrbative QCD by Muller.
Oldie but Goldie.

Paul Colby
Gold Member
one that gets small when I am far from the ensemble of charges and one that gets larger the farther away I get. I would dismiss the second solution as unphysical.
Just to expand on an already excellent answer, the solution which gets larger with increased distance is physical, it's just not the physics you were looking for. The solution which increases with distance describes sources residing at infinity. One needs both the field equations and the correct boundary conditions to describe a given physical problem.

Gold Member
It seems to me that what is unphysical may someday in the future turn to be physical, of course not including logical contradictions, which are discarded anyway.

I was and am reading quite a lot of books on GR and QFT and there's the distinction between a physical state and an unphysical state.
What is the difference?
What makes a state physical?

I know this is a bit philosophical .
Doesn't this distinction change with time and experiments?
There are many examples of unphysical or immaterial entities. Mathematics is an immaterial entity. So are the laws of logic. While certain types of thought cause the brain to “fire” differently on a scan, no neurologist on the planet can tell you what you are thinking - they must ask you for that information leading many to conclude that the soul is immaterial. Likewise, many religions teach that “God” is immaterial. You’re right that this question touches both physics and philosophy - nothing wrong with that.

vanhees71
Science Advisor
Gold Member
I don't know what your book means by "unphysical states". If it's in the context of quantum gauge-field theories, it may refer to various "ghost states", i.e., contributions from unphysical components of the gauge fields which have to be cancelled by the either unphysical Faddeev Popov ghosts.

Gold Member
I don't know what your book means by "unphysical states". If it's in the context of quantum gauge-field theories, it may refer to various "ghost states", i.e., contributions from unphysical components of the gauge fields which have to be cancelled by the either unphysical Faddeev Popov ghosts.
Why those are unphysical? do we argue that such ghosts cannot exist in reality? why not?

vanhees71
Science Advisor
Gold Member
The ghosts have been introduced in the formalism of calculating proper vertex functions to cancel the contribution from unphysical pieces of the gauge fields.

Take, as the most simple example, QED, i.e., the (un-Higgsed) U(1) gauge theory describing a Dirac field ("electrons and positrons") interacting through a minimally coupled U(1) gauge field ("photons"). First you start with four gauge fields.

The physical meaning is in the asymptotic free states, and as you well know, what's described with them are free photons being composed in terms of momentum-helicity single-photon Fock states. Of these only the 2 helicity states are physical, i.e., two field degrees of freedom are unphysical already in the sense of asymptic free states.

Now when calculating higher-order corrections (loops in Feynman diagrams) you formally sum also over the unphysical states, and this is on the first glance fatal, because it means you sum over states which have a negative norm, and the so calculated naive S-matrix wouldn't be unitary and also the microcausality condition may does not hold.

The path-integral formalism then shows that the correct way is to integrate over one physical field configuration only once and not over the infinitely many possibilities describe one and the same state expressed in different "gauges". So you choose a gauge-fixing condition and make sure by an appropriate functional ##\delta## distribution that you only pick out the one gauge configuration subject to this constraint. However, this is not so easily implemented in terms of a perturbative treatment, and that's why you use the trick to integrate over the gauge group and then expressing the arising functional determinant in the pertinent path-integral measure in terms of also unphysical Faddeev Popov ghost fields (which are formally scalar fields but implemented in the path integral as Grassmann fields, because the corresponding functional Jacobi determinant is in the numerator rather than the denominator), and these exactly cancel the contributions from the unphysical states of the gauge field.

In QED the Faddeev-Popov ghosts in the usual linear gauges are non-interacting and thus can be omitted, but that's not so for the non-Abelian case.

For details, see my QFT notes:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

hutchphd and Paul Colby